Suppose that an object is 100 lbs when it is at sea level. find the value of c that makes the equation true. (Sea level is 3963 miles from the center of the earth.) Use the value of c you found in the previous question to determin how much the object would weight in the Death valley (282 feet below sea level
You need to show the equation you are referring to, that has a constant c in it, which you are supposed to make true.
I will assume the equation is, for the 100# object in question,
W = C/R^2
where R is the diatance of the object from the center of the Earth in miles and W is in lb.
C = W R^2 = 100*3963^2 = 1.5705*10^9 lb
*mile^2
At 282 ft below sea level, the distance from the center of the earth is reduced by 282/5280 = 0.0534 miles, so the new weight W' is
W' = C/R^2
= 1.5705*10^9 lb /(3963-.0534)^2
= 100.064 lb
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To find the value of c that makes the equation true, we need to use Newton's law of universal gravitation. According to this law, the force of gravity acting on an object is given by the equation F = G * (m1 * m2) / r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects attracting each other, and r is the distance between the centers of the two objects.
In this case, the force of gravity acting on the object is equal to its weight. The weight of an object is given by the equation W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity.
First, let's convert the weight of the object from pounds to kilograms, since the gravitational constant is usually given in terms of SI units. 1 pound is approximately equal to 0.4536 kilograms. Therefore, the weight of the object at sea level is (100 lbs) * (0.4536 kg/lb) ≈ 45.36 kg.
Now, let's find the acceleration due to gravity at sea level. We can use the equation g = G * (M_e / r^2), where M_e is the mass of the Earth and r is the distance from the object to the center of the Earth. The mass of the Earth is approximately 5.972 × 10^24 kg, and the distance from sea level to the center of the Earth is 3963 miles or approximately 6.378 × 10^6 meters.
Plugging in the values, we have g = (6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg) / (6.378 × 10^6 m)^2 ≈ 9.819 m/s^2.
Now, let's find the value of c that makes the equation true. Using the formula W = m * g, we have 45.36 kg * 9.819 m/s^2 = c * 9.819 m/s^2.
Simplifying the equation, we get c = 45.36 kg.
To determine how much the object would weigh in Death Valley, which is 282 feet below sea level, we need to calculate the acceleration due to gravity at that location. The distance from Death Valley to the center of the Earth would be 3963 miles + (282 feet * 0.3048 m/ft) = approximately 6.375 × 10^6 meters.
Using the equation g = G * (M_e / r^2) and substituting the new value for r, we calculate g = (6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg) / (6.375 × 10^6 m)^2 ≈ 9.821 m/s^2.
Now we can use the weight formula W = m * g and substitute the weight value of 45.36 kg and the new value of g to find the weight of the object in Death Valley. W = 45.36 kg * 9.821 m/s^2 ≈ 445.09 N.
Therefore, the object would weigh approximately 445.09 Newtons in Death Valley, 282 feet below sea level.